Part 3 makes perfect sense to me, it's an "x = x+1" style line.
N is (((the number of digits of X), divided by 2), rounded down.)
Edit: hmm, maybe I've got it wrong, the convergence is very slow. Anyways, here's how it works out for me:
X = 105362
X = 10.5362 so N = 2 (half the number of digits to the right of the decimal point...)
Y = 2
Y = 2*(300-10.5362*2*2) = 2.578552
Y = 2.578552*(300-10.5362*2.578552*2.578552) = 2.9646326
Y = 2.9646326*(300-10.5362*2.9646326*2.9646326) = 3.0742773
It should converge to around 3.24, at which point you would shift the decimal point back 2 digits to get 324.
Yeah, it should read "repeat step three". Frequently, if you want to find x satisfying f(x) = x you can just let x be the limit of the sequence t, f(t), f(f(t)), f(f(f(t))) where t is some arbitrary number. In this case, we have:
So it looks like the algorithm calculates an inverse square root instead, and then multiplies it by the original number (step 5.)
Exactly. The nice thing about this method is that it converges nicely even if you make errors (arithmetical or rounding) along the way -- so you can start by working with only a few digits and only use full precision right at the end.
You could compute square roots using a direct NR iteration (Y := (Y + X/Y) / 2), but that requires that you perform a division at each step -- most people find multiplications to be easier and faster.
1. Move the decimal point 2N digits to the left, so that your value X is of the form a.bcdef... or ab.cdef... Remember the value N.
2. If X now has one digit before the decimal place, set Y = 5. Otherwise, set Y = 2.
3. Compute Y = Y * (300 - X * Y * Y) / 200 to as many decimal places as you want. Feel free to start by computing only a few decimal places.
4. Repeat step 2 until you're computing Y to the number of decimal places you want in your answer and it's not changing.
5. Compute Y = X * Y.
6. Move the decimal point in Y to the right by N-1 places.